Solving Scheduling Problem: Probability of Completing Projects A and B at Time t

[Jimmy Snyder]

The probability that project a will be complete at time t after it begins is given as:

$$\[ p_a(t) = \left\{ \begin{array}{ll} 0 & \mbox{if t \leq t_1} \\ \frac {(t - t_1)}{t_2-t_1} & \mbox{if t_1 \leq t \leq t_2} \\ 1 & \mbox{if t_2 \leq t} \right. \]$$

In other words, the project requires at least $t_1$ to complete and will take no more than $t_2$. The probability of completion at any time between $t_1$ and $t_2$ is a linear function of t.

The probability that project b will be complete at time t after it begins is given as:
$$\[ p_b(t) = \left\{ \begin{array}{ll} 0 & \mbox{if t \leq t_3} \\ \frac {(t - t_3)}{t_4-t_3} & \mbox{if t_3 \leq t \leq t_4} \\ 1 & \mbox{if t_4 \leq t} \right. \]$$


Project b will begin the moment project a is complete. Given t, what is the probability $p_{ab}(t)$ that both projects will be complete at time t. Obviously,
$$\[ p_{ab}(t) = \left\{ \begin{array}{ll} 0 & \mbox{if t \leq t_1 + t_3} \\ 1 & \mbox{if t_2 + t_4 \leq t} \right. \]$$.

What is $p_{ab}(t)$ for $t_1 + t_3 \leq t \leq t_2 + t_4$?

 

[Valkarie]

In this case, the probability that both projects will be completed at time t is the product of the probabilities that each project will be completed by time t: \[p_{ab}(t) = p_a(t-t_3) \cdot p_b(t-t_1)\]

 

[tacman]

To calculate p_{ab}(t) for t_1 + t_3 \leq t \leq t_2 + t_4, we need to consider the overlap between the probabilities of project a and project b completing at time t. This can be visualized as the shaded area in the graph below:

[insert graph here]

The shaded area represents the probability of both projects a and b completing at time t. It can be calculated by finding the minimum of the two probabilities at each time t within the range t_1 + t_3 \leq t \leq t_2 + t_4.

Therefore, p_{ab}(t) for t_1 + t_3 \leq t \leq t_2 + t_4 can be calculated as:

\[ p_{ab}(t) = \left\{ \begin{array}{ll}
0 & \mbox{if t \leq t_1 + t_3$} \\
\min \{p_a(t), p_b(t)\} & \mbox{if t_1 + t_3 \leq t \leq t_2 + t_4$} \\
1 & \mbox{if t_2 + t_4 \leq t}
\right. \]

In other words, p_{ab}(t) will be equal to the minimum of p_a(t) and p_b(t) within the range t_1 + t_3 \leq t \leq t_2 + t_4. This means that if either project a or project b has a lower probability of completion at time t, then the overall probability of both projects being completed at that time will be equal to that lower probability.

Overall, this approach allows us to calculate the probability of both projects being completed at any given time t, taking into account the overlap between their individual completion probabilities.